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    pdftitle={MATH235 (Mathematics IIA)},
    pdfsubject={Solution To Assignment 2},
    pdfauthor={Vafa Khalighi, SID: 41206312 <vafa@users.berlios.de>},
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\title{MATH235, Solution To Assignment 2}
\author{Vafa \underline{Khalighi}, SID:~41206312\\
\footnotesize\texttt{<vafa@users.berlios.de>}}
\begin{document}
\maketitle
\tableofcontents
\section{Question 1}
\subsection{Part a}
\begin{equation}
g(x,y)=
\begin{cases}
\frac{xy}{x^2+y^2} & \text{for all } (x,y)\not=(0,0)\\
0 & \text{for } (x,y)=(0,0)
\end{cases}\label{eq1}
\end{equation}
$\partial g/\partial x$ exits everywhere on $\mathbb{R}^2$ if 
\begin{equation}\lim_{h\to0}\frac{g(x+h,y)-g(x,y)}{h}\label{eq2}\end{equation}
exists. Now we have two conditions to check:
\begin{enumerate}[i.]
\item For all $(x,y)\not=0$, we have
\begin{align}
\lim_{h\to0}\frac{g(x+h,y)-g(x,y)}{h}&=\lim_{h\to0}\frac{\frac{(x+h)y}{(x+h)^2+y^2}-\frac{xy}{x^2+y^2}}{h}\nonumber\\
&=\lim_{h\to0}\frac{\frac{y(x+h)(x^2+y^2)-xy\left((x+h)^2+y^2\right)}{\left((x+h)^2+y^2\right)(x^2+y^2)}}{h}\nonumber\\
&=\lim_{h\to0}\frac{\frac{(xy+hy)(x^2+y^2)-xy(x^2+2hx+h^2+y^2)}{\left((x+h)^2+y^2\right)(x^2+y^2)}}{h}\nonumber\\
&=\lim_{h\to0}\frac{\frac{x^3y+xy^3+hx^2y+hy^3-x^3y-2hx^2y-h^2xy-xy^3}{\left((x+h)^2+y^2\right)(x^2+y^2)}}{h}\nonumber\\
&=\lim_{h\to0}\frac{\frac{-hx^2y+hy^3-h^2xy}{\left((x+h)^2+y^2\right)(x^2+y^2)}}{h}\nonumber\\
&=\lim_{h\to0}\frac{\frac{-hy(x^2-y^2+hx)}{\left((x+h)^2+y^2\right)(x^2+y^2)}}{h}\nonumber\\
&=-\lim_{h\to0}\frac{hy(x^2-y^2+hx)}{h\left[\left((x+h)^2+y^2\right)(x^2+y^2)\right]}\nonumber\\
&=-\lim_{h\to0}\frac{y(x^2-y^2+hx)}{\left((x+h)^2+y^2\right)(x^2+y^2)}\nonumber\\
&=-\frac{y(x^2-y^2)}{(x^2+y^2)^2}\label{eq3}
\end{align}
Equation \eqref{eq3} nicely shows that the limit in equation \eqref{eq2} exists and so $\partial g/\partial x$ exists for all $(x,y)\not=(0,0)$.
\item For $(x,y)=(0,0)$, we have
\begin{align}
\lim_{h\to0}\frac{g(x+h,y)-g(x,y)}{h}&=\lim_{h\to0}\frac{g(0+h,0)-g(0,0)}{h}\nonumber\\
&=\lim_{h\to0}\frac{g(h,0)-g(0,0)}{h}\nonumber\\
&=\lim_{h\to0}\frac{\frac{h\times0}{h^2+0^2}-0}{h}\nonumber\\
&=\lim_{h\to0}0=0\label{eq4}
\end{align}
Equation \eqref{eq4} nicely shows that the limit in equation \eqref{eq2} exists and so $\partial g/\partial x$ exists for $(x,y)=(0,0)$.
\end{enumerate}
Therefore $\partial g/\partial x$ exists  everywhere on $\mathbb{R}^2$.

$\partial g/\partial y$ exits everywhere on $\mathbb{R}^2$ if 
\begin{equation}\lim_{h\to0}\frac{g(x,y+h)-g(x,y)}{h}\label{eq5}\end{equation}
exists. Now we have two conditions to check:
\begin{enumerate}[i.]
\item For all $(x,y)\not=(0,0)$, we have
\begin{align}
\lim_{h\to0}\frac{g(x,y+h)-g(x,y)}{h}&=\lim_{h\to0}\frac{\frac{x(y+h)}{x^2+(y+h)^2}-\frac{xy}{x^2+y^2}}{h}\nonumber\\
&=\lim_{h\to0}\frac{\frac{x(y+h)(x^2+y^2)-xy\left(x^2+(y+h)^2\right)}{\left(x^2+(y+h)^2\right)(x^2+y^2)}}{h}\nonumber\\
&=\lim_{h\to0}\frac{\frac{(xy+hx)(x^2+y^2)-xy(x^2+2hy+h^2+y^2)}{\left(x^2+(y+h)^2\right)(x^2+y^2)}}{h}\nonumber\\
&=\lim_{h\to0}\frac{\frac{x^3y+xy^3+hx^3+hxy^2-x^3y-2hxy^2-h^2xy-xy^3}{\left(x^2+(y+h)^2\right)(x^2+y^2)}}{h}\nonumber\\
&=\lim_{h\to0}\frac{\frac{-hxy^2+hx^3-h^2xy}{\left(x^2+(y+h)^2\right)(x^2+y^2)}}{h}\nonumber\\
&=\lim_{h\to0}\frac{\frac{-hx(y^2-x^2+hy)}{\left(x^2+(y+h)^2\right)(x^2+y^2)}}{h}\nonumber\\
&=-\lim_{h\to0}\frac{hx(y^2-x^2+hy)}{h\left[\left(x^2+(y+h)^2\right)(x^2+y^2)\right]}\nonumber\\%
&=-\lim_{h\to0}\frac{x(y^2-x^2+hy)}{\left(x^2+(y+h)^2\right)(x^2+y^2)}\nonumber\\
&=-\frac{x(y^2-x^2)}{(x^2+y^2)^2}\label{eq6}
\end{align}
Equation \eqref{eq6} nicely shows that the limit in equation \eqref{eq5} exists and so $\partial g/\partial y$ exists for all $(x,y)\not=(0,0)$.
\item For $(x,y)=(0,0)$, we have
\begin{align}
\lim_{h\to0}\frac{g(x,y+h)-g(x,y)}{h}&=\lim_{h\to0}\frac{g(0,0+h)-g(0,0)}{h}\nonumber\\
&=\lim_{h\to0}\frac{g(0,h)-g(0,0)}{h}\nonumber\\
&=\lim_{h\to0}\frac{\frac{0\times h}{0^2+h^2}-0}{h}\nonumber\\
&=\lim_{h\to0}0=0\label{eq7}
\end{align}
Equation \eqref{eq7} nicely shows that the limit in equation \eqref{eq5} exists and so $\partial g/\partial y$ exists for $(x,y)=(0,0)$.
\end{enumerate}
Therefore $\partial g/\partial y$ exists  everywhere on $\mathbb{R}^2$.
\subsection{Part b}
$g$ has a total derivative (is differentiable) at $(0,0)$ if
\begin{equation}
\lim_{(x,y)\to(0,0)}\frac{g(x,y)-g(0,0)-\left[\frac{\partial g}{\partial x}(0,0)\right](x-0)-\left[\frac{\partial g}{\partial y}(0,0)\right](y-0)}{\|(x,y)-(0,0)\|}=0\label{eq8}
\end{equation}
We know from equation \eqref{eq1} that 
\begin{equation}
g(0,0)=0\label{eq9}
\end{equation}
We know from equation \eqref{eq4} that
\begin{equation}
\frac{\partial g}{\partial x}(0,0)=0\label{eq10}
\end{equation}
We know from equation \eqref{eq7} that
\begin{equation}
\frac{\partial g}{\partial y}(0,0)=0\label{eq11}
\end{equation}
By substituting equations \eqref{eq9}, \eqref{eq10} and \eqref{eq11} into equation \eqref{eq8}, we have
\begin{align}
\lim_{(x,y)\to(0,0)}\frac{\frac{xy}{x^2+y^2}-0-0(x-0)-0(y-0)}{\sqrt{x^2+y^2}}&=\lim_{(x,y)\to(0,0)}\frac{\frac{xy}{x^2+y^2}}{\sqrt{x^2+y^2}}\nonumber\\
&=\lim_{(x,y)\to(0,0)}\frac{xy}{(x^2+y^2)^{3/2}}\label{eq12}
\end{align}
Now our aim to see if the limit of equation \eqref{eq12} exist. Let $f(x,y)=(xy)/(x^2+y^2)^{3/2}$. Along the $x$-axis, we have $y=0$, and so $f(x,y)=0$ and then
\[\lim_{(x,y)\to(0,0)}\frac{xy}{(x^2+y^2)^{3/2}}=0\quad(\text{along the } x\text{-axis})\]
Along the $y$-axis, we have $x=0$, and so $f(x,y)=0$ and then
\[\lim_{(x,y)\to(0,0)}\frac{xy}{(x^2+y^2)^{3/2}}=0\quad(\text{along the } y\text{-axis})\]
Along the line $y=x$, we have $f(x,y)=x^2/(2x^2)^{3/2}$, and so
\begin{align}
\lim_{(x,y)\to(0,0)}\frac{xy}{(x^2+y^2)^{3/2}}&=\lim_{(x,y)\to(0,0)}\frac{x^2}{2\sqrt{2}x^3}\nonumber\\
&=\frac{1}{2\sqrt{2}}\lim_{(x,y)\to(0,0)}\frac{1}{x}\quad (\text{along the line } y=x)\label{eq13}
\end{align}
The limit in equation \eqref{eq13} does not exist because
\[\lim_{x\to0^+}\frac{1}{x}=\infty\quad\text{and}\quad\lim_{x\to0^-}\frac{1}{x}=-\infty\]
Now since we have obtained different limits along different paths, then the limit in equation \eqref{eq12} and consequently the limit in equation \eqref{eq8} do not exist and hence $g(x,y)$ is not differentiable (does not have a total derivative) at the point $(0,0)$.
\subsection{Part c}
$g$ is continuous at $(0,0)$ if 
\[\lim_{(x,y)\to(0,0)}g(x,y)=g(0,0)=0\]
Now 
\begin{equation}
\lim_{(x,y)\to(0,0)}g(x,y)=\lim_{(x,y)\to(0,0)}\frac{xy}{x^2+y^2}\label{eq14}
\end{equation}
Let $h(x,y)=xy/(x^2+y^2)$. Along the $x$-axis, we have $y=0$ and so $h(x,y)=0$, therefore
\[\lim_{(x,y)\to(0,0)}\frac{xy}{x^2+y^2}=0\quad (\text{along the }x\text{-axis})\]
Along the $y$-axis, we have $x=0$ and so $h(x,y)=0$, therefore
\[\lim_{(x,y)\to(0,0)}\frac{xy}{x^2+y^2}=0\quad (\text{along the }y\text{-axis})\]
Along the line $y=x$, we have $h(x,y)=x^2/2x^2=1/2$, and therefore
\[\lim_{(x,y)\to(0,0)}\frac{xy}{x^2+y^2}=\frac{1}{2}\quad (\text{along the line }y=x)\]
Since we have obtained different limits along different paths, then the given limit in equation \eqref{eq14} does not exist and hence the function $g$ is not continuous at $(0,0)$.
\section{Question 2}
\subsection{Part a}
\[\nabla f(x,y,z)=\left(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z}\right)=(2x,2y,2z)\]
Let $\mathbf{u}=(1,-1,0)$. $\mathbf{u}$ is not a unit vector so let
\[\mathbf{v}=\frac{\mathbf{u}}{\|\mathbf{u}\|}=\left(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}},0\right)\]
The directional derivative is
\[\nabla f\cdot \mathbf{v}=(2x,2y,2z)\cdot\left(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}},0\right)\]
which, at the point $(1,1,1)$, becomes
\[(2,2,2)\cdot\left(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}},0\right)=\sqrt{2}-\sqrt{2}+0=0\]
\subsection{Part b}
Let $z=2x^2+y^2$ and let $f(x,y,z)=2x^2+y^2-z=0$, then
\begin{align*}
\nabla f(x,y,z)&=\left(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z}\right)=(4x,2y,-1)\\
\nabla f(2,1,9)&=(8,2,-1)
\end{align*}
The equation of the tangent plane is
\begin{align*}
\nabla f(2,1,9)\cdot(x-2,y-1,z-9)=0&\Rightarrow (8,2,-1)\cdot(x-2,y-1,z-9)=0\\
&\Rightarrow 8(x-2)+2(y-1)-(z-9)=0\\
&\Rightarrow 8x-16+2y-2-z+9=0\\
&\Rightarrow \fbox{$8x+2y-z=9$}
\end{align*}
\section{Question 3}
\subsection{Part a}
\begin{align}
\left(\frac{\partial F}{\partial x}\right)^2&=\left(\frac{1}{2}(2x)(x^2+y^2+z^2)^{-1/2}u'\left(\sqrt{x^2+y^2+z^2}\right)\right)^2\nonumber\\
&=\left(\frac{xu'\left(\sqrt{x^2+y^2+z^2}\right)}{\sqrt{x^2+y^2+z^2}}\right)^2\nonumber\\
&=\frac{x^2\left[u'\left(\sqrt{x^2+y^2+z^2}\right)\right]^2}{x^2+y^2+z^2}\label{eq15}
\end{align}
\begin{align}
\left(\frac{\partial F}{\partial y}\right)^2&=\left(\frac{1}{2}(2y)(x^2+y^2+z^2)^{-1/2}u'\left(\sqrt{x^2+y^2+z^2}\right)\right)^2\nonumber\\
&=\left(\frac{yu'\left(\sqrt{x^2+y^2+z^2}\right)}{\sqrt{x^2+y^2+z^2}}\right)^2\nonumber\\
&=\frac{y^2\left[u'\left(\sqrt{x^2+y^2+z^2}\right)\right]^2}{x^2+y^2+z^2}\label{eq16}
\end{align}
\begin{align}
\left(\frac{\partial F}{\partial z}\right)^2&=\left(\frac{1}{2}(2z)(x^2+y^2+z^2)^{-1/2}u'\left(\sqrt{x^2+y^2+z^2}\right)\right)^2\nonumber\\
&=\left(\frac{zu'\left(\sqrt{x^2+y^2+z^2}\right)}{\sqrt{x^2+y^2+z^2}}\right)^2\nonumber\\
&=\frac{z^2\left[u'\left(\sqrt{x^2+y^2+z^2}\right)\right]^2}{x^2+y^2+z^2}\label{eq17}
\end{align}
By adding equations \eqref{eq15}, \eqref{eq16} and \eqref{eq17}, we have
\begin{align*}
\left(\frac{\partial F}{\partial x}\right)^2+\left(\frac{\partial F}{\partial y}\right)^2+\left(\frac{\partial F}{\partial z}\right)^2&=\frac{x^2\left[u'\left(\sqrt{x^2+y^2+z^2}\right)\right]^2}{x^2+y^2+z^2}+\frac{y^2\left[u'\left(\sqrt{x^2+y^2+z^2}\right)\right]^2}{x^2+y^2+z^2}\\
&\qquad+\frac{z^2\left[u'\left(\sqrt{x^2+y^2+z^2}\right)\right]^2}{x^2+y^2+z^2}\\
&=\frac{\left[u'\left(\sqrt{x^2+y^2+z^2}\right)\right]^2}{x^2+y^2+z^2}\left(x^2+y^2+z^2\right)\\
&=\left[u'\left(\sqrt{x^2+y^2+z^2}\right)\right]^2
\end{align*}
\subsection{Part b}
\begin{align*}
\mathbf{DG(x)}&=
\begin{bmatrix}
\frac{\partial}{\partial x}(xy-z^2)&\frac{\partial}{\partial y}(xy-z^2)&\frac{\partial}{\partial z}(xy-z^2)\\
\frac{\partial }{\partial x}(zy+x^3)&\frac{\partial }{\partial y}(zy+x^3)&\frac{\partial }{\partial z}(zy+x^3)
\end{bmatrix}\\
&=\begin{bmatrix}
y&x&-2z\\
3x^2&z&y
\end{bmatrix}
\end{align*}

















\section{Question 5}
\subsection{Part a}
Let $\mathbf{P}=c_0+c_1x+c_2x^2+c_3x^3$ be a polynomial in $\mathbb{P}_3$ (domain), and define $D:\mathbb{P}_3\to\mathbb{P}_3$ by
\begin{align*}
D(\mathbf{P})=D(P(x))=P''(x)&=\left(c_0+c_1x+c_2x^2+c_3x^3\right)''\\
&=\left(\left(c_0+c_1x+c_2x^2+c_3x^3\right)'\right)'\\
&=\left(c_1+2c_2x+3c_3x^2\right)'\\
&=2c_2+6c_3x
\end{align*}
$D$ is a linear transformation, since for any scalar $k$ and any polynomials $\mathbf{P}_1$ and $\mathbf{P}_2$ in $\mathbb{P}_3$ (domain), we have
\begin{align*}
D(\mathbf{P}_1+\mathbf{P}_2)=D(P_1(x)+P_2(x))&=\left(P_1(x)+P_2(x)\right)''\\
&=\left(\left(P_1(x)+P_2(x)\right)'\right)'\\
&=\left(P'_1(x)+P'_2(x)\right)'\\
&=P''_1(x)+P''_2(x)\\
&=D(\mathbf{P}_1)+D(\mathbf{P}_2)
\intertext{and}
D(K\mathbf{P})=D(KP(x))=(KP(x))''&=\left((KP(x)'\right)'=\left(K(P(x))'\right)'\\
&=\left(KP'(x)\right)'=K\left(P'(x)\right)'\\
&=KP''(x)=KD(\mathbf{P})
\end{align*}
\subsection{Part b}
\begin{center}
\begin{tikzpicture}
\node [notebox] (box){%
    \begin{minipage}{0.7\textwidth}
We use the notation ``Ker$(D)$" instead ``$N(D)$'' here.
    \end{minipage}
};
\node[notetitle, right=10pt] at (box.north west) {Note};
\end{tikzpicture}\label{ker}
\end{center}
The kernel of $D$ is the set of polynomials in $\mathbb{P}_3$ (domain) with the second derivative 0. From calculus, we know that 
\begin{align*}\text{Ker}(D)&=\left\{\text{constant polynomails }(\text{i.e. }P(x)=c_0),\right.\\
&\left.\qquad \text{ polynomials of degree $1$ } (\text{i.e. }P(x)=c_0+c_1x)\right\}
\end{align*}
\subsection{Part c}
The Range of $D$ is the set of polynomials in $\mathbb{P}_3$ (codomain) that are images under $D$ of at least one polynomial in $\mathbb{P}_3$ (domain), hence we have
\[\text{R}(D)=\left\{P(x)=2c_2, P(x)=2c_2+6c_3x\right\}\]
\subsection{Part d}
No, because if $\mathbf{P}=c_0+c_1x+c_2x^2+c_3x^3$ and $\mathbf{Q}=d_0+d_1x+d_2x^2+d_3x^3$ are distinct polynomials, then they differ in at least one coefficient. So, let $c_1=d_1, c_2=d_2$ and $c_3=d_3$ but $c_0\not=d_0$. Now we have
\[\mathbf{P}=P(x)=c_0+c_1x+c_2x^2+c^3x^3\quad\text{and}\quad \mathbf{Q}=Q(x)=d_0+c_1x+c_2x^2+c^3x^3\]
but 
\begin{equation}
\begin{split}
D(\mathbf{P})=D(P(x))=P''(x)&=\left(c_0+c_1x+c_2x^2+c_3x^3\right)''\\
&=\left(\left(c_0+c_1x+c_2x^2+c_3x^3\right)'\right)'\\
&=\left(c_1+2c_2x+3c_3x^2\right)'\\
&=2c_2+6c_3x
\end{split}\label{eq18}
\end{equation}
and
\begin{equation}
\begin{split}
D(\mathbf{Q})=D(Q(x))=Q''(x)&=\left(d_0+c_1x+c_2x^2+c_3x^3\right)''\\
&=\left(\left(d_0+c_1x+c_2x^2+c_3x^3\right)'\right)'\\
&=\left(c_1+2c_2x+3c_3x^2\right)'\\
&=2c_2+6c_3x
\end{split}\label{eq19}
\end{equation}
Now from equations \eqref{eq18} and \eqref{eq19}, we can clearly see that
\[D(\mathbf{P})=D(\mathbf{Q})=2c_2+6c_3x\]
So this linear transformation is not one-to-one (injective) because it maps polynomials that differ by a constant into the same polynomial. For example, $D(x^2)=D(x^2+9)=2$.
\section{Question 6}
\subsection{Part a}
\begin{align*}
S_x&=x-2y&T_x&=0x+y&&\\
S_y&=3x+y&T_y&=x+0y&&
\intertext{Therefore}
[S]&=\begin{bmatrix}
1&-2\\
3&1
\end{bmatrix}
&[T]&=\begin{bmatrix}
0&1\\
1&0
\end{bmatrix}
&&
\end{align*}
\subsection{Part b}
\begin{align*}
T^\circ S=T^\circ\left(S(x,y)\right)&=T\left(x-2y,3x+y\right)\\
&=\left(3x+y,x-2y\right)
\intertext{So}
\left(T^\circ S\right)_x&=3x+y\\
\left(T^\circ S\right)_y&=x-2y
\intertext{Therefore}
\left[T^\circ S\right]&=
\begin{bmatrix}
3&1\\
1&-2
\end{bmatrix}
\end{align*}
\subsection{Part c}
\begin{align*}
[T^\circ S]=[T][S]&=\begin{bmatrix}
0&1\\
1&0
\end{bmatrix}
\begin{bmatrix}
1&-2\\
3&1
\end{bmatrix}\\
&=\begin{bmatrix}
(0\times 1)+(1\times 3)&(0\times -2)+(1\times 1)\\
(1\times 1)+(0\times 3)&(1\times -2)+(0\times 1)
\end{bmatrix}\\
&=\begin{bmatrix}
3&1\\
1&-2
\end{bmatrix}
\end{align*}
\section{Question 7}
\begin{center}
\begin{tikzpicture}
\node [notebox] (box){%
    \begin{minipage}{0.8\textwidth}
We use the notation ``Ker$(T)$" instead ``$N(T)$'' as we have used it in section \ref{ker}.
    \end{minipage}
};
\node[notetitle, right=10pt] at (box.north west) {Note};
\end{tikzpicture}%
\end{center}
Let $V$ and $W$ be vector sapces. If $p$ states that ``a linear transformation $T:V\to W$ is injective'' and $q$ states that ``$\text{Ker}(T)=\{\mathbf{0}\}$'', then we are required to show that ``$p\Leftrightarrow q$''.
\begin{description}
\item[Show $\mathbf{p}\boldsymbol{\Rightarrow}\mathbf{q}$: ]Assume that $T$ is injective and let $\mathbf{v}$ be any vector in $\text{Ker}(T)$. Since $\mathbf{v}$ and $\mathbf{0}$ both lie in $\text{Ker}(T)$, we have $T(\mathbf{v})=\mathbf{0}$ and $T(\mathbf{0})=\mathbf{0}$, so $T(\mathbf{v})=T(\mathbf{0})$. But this implies that $\mathbf{v}=\mathbf{0}$, since $T$ is injective; thus $\text{Ker}(T)$ only contains the zero vector.
\item[Show $\mathbf{q}\boldsymbol{\Rightarrow}\mathbf{p}$: ]Assume that $\text{Ker}(T)=\{\mathbf{0}\}$ and that $\mathbf{v}$ and $\mathbf{w}$ are distinct vectors in $V$; that is
\begin{equation}
\mathbf{v}-\mathbf{w}\not=\mathbf{0}\label{eq20}
\end{equation}
To prove that $T$ is injective, we must show that $T(\mathbf{v})$ and $T(\mathbf{w})$ are distinct vectors, but if $T(\mathbf{v})$ and $T(\mathbf{w})$ were not distinct vectors, then we would have $T(\mathbf{v})=T(\mathbf{w})$. Therefore
\begin{align}
T(\mathbf{v})=T(\mathbf{w})&\Rightarrow T(\mathbf{v})-T(\mathbf{w})=\mathbf{0}&&\nonumber\\
&\Rightarrow T(\mathbf{v})+(-T(\mathbf{w}))=\mathbf{0}&&\nonumber\\
&\Rightarrow T(\mathbf{v})+(-1)T(\mathbf{w})=\mathbf{0}&&\nonumber\\
&\Rightarrow T(\mathbf{v})+T((-1)\mathbf{w})=\mathbf{0}&&(\text{linear transformation implies }\nonumber\\
&&&T(k\mathbf{u})=kT(\mathbf{u}))&&\nonumber\\
&\Rightarrow T(\mathbf{v})+T(-\mathbf{w})=\mathbf{0}&&\nonumber\\
&\Rightarrow \fbox{$T(\mathbf{v}-\mathbf{w})=\mathbf{0}$}&&(\text{linear transformation implies }\nonumber\\
&&&\qquad T(\mathbf{s})+T(\mathbf{u})=T(\mathbf{s}+\mathbf{u}))\nonumber
\end{align}
and so $\mathbf{v}-\mathbf{w}$ is in the kernel of $T$. Since $\text{Ker}(T)=\{\mathbf{0}\}$, this implies that $\mathbf{v}-\mathbf{w}=\mathbf{0}$ which contradicts equation \eqref{eq20}. Thus $T(\mathbf{v})$ and $T(\mathbf{w})$ must be distinct.
\end{description}
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